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I was puzzled at first, over the fact that we define the group-action of the symmetric group $S_k$ on $k$-tensors $$z \in \mathcal{T}^k(M)$$ as $$\sigma z = \sigma(m_1 \otimes \cdots \otimes m_k) = (m_{\sigma^{-1}(1)} \otimes \cdots \otimes m_{\sigma^{-1}(k)}).$$

After reading exercise $5.1.8$ in D & F, I believe I understand it better. But I am still not sure why we can´t define it as $$\sigma z = \sigma (m_1 \otimes \cdots \otimes m_k) = (m_{\sigma(1)} \otimes \cdots \otimes m_{\sigma(k)}).$$

I believe the axiom $$(\sigma_1 \cdot \sigma_2) \cdot z = \sigma_1 \cdot (\sigma_2 \cdot z)$$ for group-action fails, but why does it fail, exactly? According to the authors, if my memory serves me right, it becomes a right group-action.

I would be happy to receive clarification.

Ben123
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    Yes, it's a right action, just check it carefully. The phenomenon here is exactly the same reason why to shift the graph of a function $f(x)$ in the positive direction by $c$ you have to subtract $c$ to get $f(x - c)$ rather than add it. – Qiaochu Yuan Oct 11 '23 at 03:37

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